%% Math 131, Fall, 2003
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\begin{document}

\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Solutions to Worksheet for Thursday, October~2
\end{center}

\be

\item The difference quotient
  \[
  \frac{f(x+h)-f(x-h)}{2h}
  \]
  can be used to estimate derivatives numerically, and for a given
  value of~$h$ is more accurate than the difference quotient used to
  define the derivative. (Theoretically, the two expressions give the
  same answer when $f$~is differentiable.) Compute the difference
  quotient above for the following functions:
  \[
  f(x)=x^2,\qquad g(x)=\frac{1}{x}.
  \]

  \soln Since $(x+h)^2-(x-h)^2=(x^2+2xh+h^2)-(x^2-2xh+h^2)=4xh$, the
  symmetric difference quotient for~$f$ is $2x$, which is already the
  exact value of the derivative. For~$g$, the symmetric difference
  quotient is
  \[
  \frac{1}{2h}\left(\frac{1}{x+h}-\frac{1}{x-h}\right)
  =\frac{1}{2h}\left(\frac{(x-h)-(x+h)}{(x+h)(x-h)}\right)
  =-\frac{1}{x^2-h^2}.
  \]

\item Plot the data in the table below on the grid provided. Use the
  ordinary difference quotient to estimate the value of~$f'(x)$ for
  $x=0.7$, $0.9$, $1.0$, and~$1.1$. Then use the difference quotient
  from the previous problem to estimate~$f'(1)$. Which answer best
  fits the data?

  \[
  \begin{array}{|c||r|r|r|r|r|r|}
    \hline
    \vp x & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 \\
    \hline
    \vp f(x) & -1.52 & -1.72 & -1.88 & -2.00 & -2.08 & -2.12 \\
    \hline
    \end{array}
  \]

  \bigskip

  \soln Plugging in numbers gives
  \[
  f'(0.7)\simeq\frac{f(0.8)-f(0.7)}{0.8-0.7}
  = \frac{-0.2}{0.1}=-2,\quad
  f'(0.9)\simeq -1.6,\quad
  f'(1.0)\simeq-0.8,\quad
  f'(1.1)\simeq-0.4.
  \]
  The symmetric difference quotient at $x=1$ gives $f'(1)\simeq-1$,
  which fits the data better than the answer found above.

\begin{center}
  \input{hsoln03a.eepic}
\end{center}

\vfil\eject

\item The graph of~$f$ is shown below. On the same grid, sketch the
  graph of~$f'$; try to be as accurate as possible with the vertical
  scale. 

\begin{center}
  \input{hsoln03b.eepic}
\end{center}

\vfil

\item The graph of~$f'$ is shown below. On the same grid, sketch the
  graph of~$f$, assuming that $f(0)=0$. As above, strive for accuracy
  with the vertical scale.

\begin{center}
  \input{hsoln03c.eepic}
\end{center}

\ee

\vfil\eject

\end{document}
